Question : A parallel plate capacitor with air between the plates has a capacitance of 8µF. What will be the capacitance if the distance between the plates is doubled and the space between them is filled with a substance of dielectric constant K=2?
Doubt by Ananya
Solution :
Case I
Area of each plate = A
Separation between the two plates = d
Permittivity of free space = ε₀
Capacitance = C₀=ε₀A/d=8µF
Case II
A'=A
d=2d
Permittivity of the medium = ε=Kε₀=2ε₀
New Capacitance
C=εA'/d'
C=εA'/d'
C=Kε₀A/(2d)
C=2ε₀A/(2d)
C=ε₀A/d
C=C₀
C= 8µF.
C=ε₀A/d
C=C₀
C= 8µF.
Hence, the new capacitance will remains equal to the initial capacitance of 8µF.
Similar Question :
A parallel plate capacitor with air between the plates has a capacitance of 8 µF. What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?
Click Here for the Answer
96µF
